Recall a function $f:[-a,a] \rightarrow R$ is said to be even if $f(x)=f(-x)$.

Let $f$ be an integrable, even function.

Prove that: $ \int_{-a}^a f\,$ = $2 \int_0^a f\,$.


I'm trying to prove this using Darboux/Riemann Integrals, as stated in the title.

I'm unsure how to interpret $f(x)=f(-x)$. My best guess was to say that $U(f,P)=U(f,-P)$ after I define a partition but, even if that's right, I can't see how that works.

My proof so far is as follows:

As $f$ is integrable, $\exists$ a sequence of partitions $P_n$ on $[-a,a]$ : $\lim\limits_{n \to \infty} (U(f,P_n)-L(f,P_n))=0$ and $\lim\limits_{n \to \infty} (U(f,P_n))=$ $\int_{-a}^a f\,$.

Now, let ${Q_n}$ be a refinement of $P_n$ such that $Q_n=P_n \cup (-P_n)$. Then, $Q_n$ is symmetric about the origin and:

(i.) $\lim\limits_{n \to \infty} (U(f,Q_n)-L(f,Q_n)=0$ (see ***).

(ii.) $L(f,P_n) \le L(f,Q_n) \le U(f,Q_n) \le U(f,P_n)$, by refinement.

(iii.) $\lim\limits_{n \to \infty} (U(f,Q_n))=$ $\int_{-a}^a f\,$, by (i.).

***$\lim\limits_{n \to \infty} (U(f,Q_n)-L(f,Q_n))=$ $\lim\limits_{n \to \infty} ([U(f,P_n)+U(f,-P_n)]-[L(f,P_n)+L(f,-P_n)])=0$

Here is where I get stuck...

As $f$ is even, $U(f,P_n)=U(f,-P_n)$ and $L(f,P_n)=L(f,-P_n)$.

And that's it.. I believe the final lines look like this:

Thus, $U(f,Q_n)=U(f,P_n \cup (-P_n))=U(f,P_n)+U(f,-P_n)=U(f,-P_n)+U(f,-P_n)=2U(f,-P_n)$.

And, as $ \int_{-a}^a f\,$ = $\int_{-a}^0 f\,$ $+$ $\int_0^a f\,$,

$\lim\limits_{n \to \infty} (2U(f,-P_n)=$ $\lim\limits_{n \to \infty} (U(f,Q_n)$

Hence, $ \int_{-a}^a f\,$ $=$ $\int_{-a}^0 f\,$ $+$ $\int_0^a f\,$ $=$ $\int_0^a f\,$ $+$ $\int_0^a f\,$ $=$ $2\int_0^a f\,$.

End Proof
Any help would be extremely helpful. As you can see, I get pretty lost at the end there. Thank you.

  • $\begingroup$ I edited the last line "Hence..." because I forgot a negative sign on one of the limits. Also, I may have to include the set {0} in the union, but I am not sure. So it may have to look like this: $Q_n = P_n \cup (-P_n) \cup$ {$0$}. $\endgroup$ – Marcus Apr 2 '18 at 19:22

Since $f$ is Riemann integrable on $[-a,a]$, by the Riemann criterion, for any $\epsilon > 0$ there exists a partition $P'$ of $[-a,a]$ such that $U(f,P') - L(f,P') < \epsilon.$

If the point $x=0$ is not included in $P'$, we can add it to obtain a partition $P$ which refines $P'$. Otherwise take $P = P'$ when it includes $x = 0$. Since $P' \subset P$ we have $$L(f,P') \leqslant L(f,P) \leqslant U(f,P) \leqslant U(f,P')$$ and $$U(f,P) - L(f,P) \leqslant U(f,P') - L(f,P') < \epsilon.$$

Let us denote the points in $P$ as

$$-a = y_0 < y_1 < \ldots < y_n = 0 = x_0 < x_1< \ldots x_m = a,$$

where we take into consideration the fact that there may be a different number of points in the partition below and above $0$. Note that $P^- = (y_0,y_1, \ldots,y_n)$ is a partition of $[-a,0]$ and $P^+ = (x_0,x_1, \ldots,x_n)$ is a partition of $[0,a]$.

We can write

$$U(f,P) = U(f,P^-) + U(f,P^+), \,\,\, L(f,P) = L(f,P^-) + L(f,P^+), $$

and it follows that

$$U(f,P^-) - L(f,P^-) + U(f,P^+) - U(f,P^+) = U(f,P) - L(f,P) < \epsilon.$$

Hence, since the upper sum minus the lower sum is positive, we have

$$U(f,P^+) - L(f,P^+) < \epsilon,$$

and the integral of $f$ over $[0,a]$ exists and is squeezed between the lower and upper sums as

$$\tag{1}L(f,P^+) \leqslant \int_0^a f \leqslant U(f,P^+).$$

Since $f(x) = f(-x)$, we have by the definition of upper and lower sums

$$\tag{2}U(f,P^-) = \sum_{j=1}^n \sup_{x \in [y_{j-1},y_j]}f(x) \,(y_j- y_{j-1}) = \sum_{j=1}^n \sup_{-x \in [-y_j,-y_{j-1}]}f(-x) \,(-y_{j-1}- (-y_j)), \\ L(f,P^-) = \sum_{j=1}^n \inf_{x \in [y_{j-1},y_j]}f(x) \,(y_j- y_{j-1}) = \sum_{j=1}^n \inf_{-x \in [-y_j,-y_{j-1}]}f(-x) \,(-y_{j-1}- (-y_j)) $$

Notice that the sums appearing on the right-hand sides of the two equations in (2) are themselves upper and lower sums with respect to a partition of $[0,a]$ since $-y_j \in [0,a]$.


$$\tag{3}L(f,P^-) \leqslant \int_0^a f \leqslant U(f,P^-).$$

Adding (1) and (3) we get

$$\tag{4}L(f,P) \leqslant 2\int_0^a f \leqslant U(P,f).$$

But $f$ is integrable over $[-a,a]$ and we must also have

$$\tag{4}L(f,P) \leqslant \int_{-a}^a f \leqslant U(P,f).$$

Thus, for any $\epsilon > 0$,

$$\left|\int_{-a}^af - 2\int_0^a f \right| < U(f,P) - L(f,P) < \epsilon,$$

and it follows that

$$\int_{-a}^af = 2\int_0^a f.$$

  • $\begingroup$ Wow, that makes so much sense. Thank you so much! $\endgroup$ – Marcus Apr 4 '18 at 23:59
  • $\begingroup$ @Marcus: You're welcome. Glad to help. $\endgroup$ – RRL Apr 5 '18 at 3:00

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